在cakephp3中的matching或innerJoinWith中传递动态变量 [英] Pass dynamic variable in matching or innerJoinWith in cakephp3

问题描述

我正在将静态值传递给以下查询，因此效果很好.如果我将其动态更改，则不会更改.

I am passing static value to my following query, then it works well. If I change it dynamic, then it doesn't.

 //works well
        $users = $this->Users->find('all')
                     ->where(['Users.role' => $role])
                     ->innerJoinWith('UserDetail',function($query){
                     return $query->where(['UserDetail.state' => "chandigarh"]); 
                   })->toArray();

       // don't work
         $state="chandigarh";
         $users = $this->Users->find('all')
                     ->where(['Users.role' => $role])
                     ->innerJoinWith('UserDetail',function($query){
                     return $query->where(['UserDetail.state' => "$state"]); 
                   })->toArray();

如何将动态的$ state变量传递给它?

How can I pass my $state variable dynamic to this?

推荐答案

这是PHP的工作方式，它实际上不是CakePHP相关的问题

This is how PHP works, it's not really a CakePHP related problem

该变量在匿名函数$ query的范围内不可用请在此处查看示例3: http://php.net/manual/zh/functions.anonymous.php

That variable is not available inside the scope of the anonymous function $query see example 3 here: http://php.net/manual/en/functions.anonymous.php

您应该使用USE结构从父作用域继承变量，如下所示:

You should inherit the variable from the parent scope by using the USE contruct like this:

$state="chandigarh";
$users = $this->Users->find('all')
            ->where(['Users.role' => $role])
            ->innerJoinWith('UserDetail',function($query) use ($state) {
                  return $query->where(['UserDetail.state' => $state]); 
            })->toArray();

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